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gtnhenbr [62]
2 years ago
6

The scatter plot below displays the elevation and mean number of clear days per year of 14 U.S. cities. Two lines

Mathematics
1 answer:
Novay_Z [31]2 years ago
4 0

Answer:

=

Step-by-step explanation:

Students use the least squares regression line to make predictions. ... Example 1 (2 minutes): Using a Line to Describe a Relationship ... the model given in the student lesson to predict a man's height using a shoe length of ... Below is a scatter plot of the data with two linear models, ... in Exercise 3 that represents this man

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The diagram shows a cuboid.<br>4 cm<br>5 cm<br>9 cm<br>What is the surface area of the cuboid?​
Salsk061 [2.6K]

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69cm ^2

Step-by-step explanation:

4 0
2 years ago
H(x) = {(3,-5), (5.-7), (6,-9), (10, -12), (12,1-16)}<br> Which of the following gives h^-1(x)
Lemur [1.5K]

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its b

Step-by-step explanation:

because each pair is inversed

7 0
3 years ago
Simplify: <img src="https://tex.z-dn.net/?f=4%5Csqrt%7B320%7D%20-%203%5Csqrt%7B80%7D" id="TexFormula1" title="4\sqrt{320} - 3\sq
aleksandrvk [35]

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20 √ 5

Step-by-step explanation:

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3 years ago
On the day of his 18th birthday harry decided to start saving up money regularly starting on that day he could save £30 on the s
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Step-by-step explanation:

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6 0
2 years ago
1. cot x sec4x = cot x + 2 tan x + tan3x
Mars2501 [29]
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
    cot(x)sec⁴(x)            cot(x)sec⁴(x)
                   0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
                   0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
                   0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
                   0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
                   0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
                   0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
                   0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
                   0 = cos⁴(x)(1 + tan²(x))²
                   0 = cos⁴(x)        or         0 = (1 + tan²(x))²
                ⁴√0 = ⁴√cos⁴(x)      or      √0 = (√1 + tan²(x))²
                   0 = cos(x)         or         0 = 1 + tan²(x)
         cos⁻¹(0) = cos⁻¹(cos(x))    or   -1 = tan²(x)
                 90 = x           or            √-1 = √tan²(x)
                                                         i = tan(x)
                                                      (No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
              sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
   sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
                               sin²(x) - cos²(x) = sin²(x) - cos²(x)
                                         + cos²(x)              + cos²(x)
                                             sin²(x) = sin²(x)
                                           - sin²(x)  - sin²(x)
                                                     0 = 0

3. 1 + sec²(x)sin²(x) = sec²(x)
           sec²(x)             sec²(x)
      cos²(x) + sin²(x) = 1
                    cos²(x) = 1 - sin²(x)
                  √cos²(x) = √(1 - sin²(x))
                     cos(x) = √(1 - sin²(x))
               cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
                                 x = 0

4. -tan²(x) + sec²(x) = 1
               -1               -1
      tan²(x) - sec²(x) = -1
                    tan²(x) = -1 + sec²
                  √tan²(x) = √(-1 + sec²(x))
                     tan(x) = √(-1 + sec²(x))
            tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
                             x = 0
5 0
3 years ago
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